Convex set contains line segment between that connects its interior and closure is contained in its interior

Question

In this question, one of the answer (by Dimitris) uses the following lemma.

Lemma. Let $A\subset \mathbb{R}^N$ be a convex set. Suppose $\text{Int}A \ne \emptyset$. If $x \in \text{Int} A$ and $y \in \text{Cl}A$, then $[x, y) \subset \text{Int}A$, where $$ [x,y) := \{y+\lambda(x-y)\mid\lambda\in(0,1]\}. $$

(I changed the statement in the case of $\mathbb{R}^N$ since it suffices for my purpose.)

I am having trouble in proving this lemma. Here is my attempt.

Pick any $x \in \text{Int} A$ and $y \in \text{Cl}A$. If $y \in \text{Int}A$, then the result follows because $A$ is convex and thus $\text{Int}A$ is as well. Suppose $y \in \text{Bdry}A:=\text{Cl}A \setminus \text{Int}A$. Pick any $z \in [x,y)$. Then, there exists $\lambda\in(0,1]$ such that $z = y + \lambda(x-y)$. If $z \notin \text{Int}A$, then $z\in \text{Bdry}A$ since $z \in \text{Cl}A$ by the convexity of $A$ and $\text{Cl} A$.

Although it is intuitively clear that $z,y\in\text{Bdry}A$ is a contradiction, how can I prove it?

Answer 1

Case 1: assume also $y\in A$. Since $x \in \text{Int} A$, there exists $\rho>0$ such that $B_\rho(x)\subset \text{Int} A$. Let $z = (1-\lambda) y + \lambda x$, $\lambda\in (0,1)$. Then $B_{\lambda\rho}(z) = z + B_{\lambda\rho}(0) = (1-\lambda)y + \lambda B_\rho(x) \subset A$, so that $z$ is an interior point of $A$.

Case 2: assume $y\in \text{Cl} A$. The ball $B_{\rho\lambda/(1-\lambda)}(y)$ contains a point $a\in A$; in particular, $a = \frac{\lambda}{1-\lambda} u + y$, for some $u\in B_\rho$. Hence, $z = (1-\lambda) a + \lambda(x-u)\in A$. By case 1, we conclude that $z\in \text{Int} A$.

Answer 2

I'll assume $N\geq 2$ (the case $N=1$ is a lot simpler). Let $r>0$ be such that the closed $r$ ball around $x$ is in $A$. By rotating and shifting, I may assume that $x$ is the origin, and that $y$ lies on the second axis.

Let $z_n$ be a sequence of points in $A$ approaching $y$. We know $p_1:=(r,0,...0)$ and $p_2:=(-r,0,...0)$ are both in $A$, so, by convexity, the triangle $\triangle_n$ with vertices $z_n,p_1,p_2$ lies entirely inside $A$.

Now, let $p$ be a point on $[x,y)$. I think it shouldn't be hard to show that for large enough $n$, the point $p$ must be in the interior of $\triangle_n$, which means it must be in the interior of $A$.